Normalized wavelets i am studying a class of functions called ridgelets, i am looking to prove that they are normalized.
Let $\varphi$ be a Real smooth function with sufficient decay and vanishing mean such that $$\int_{-\infty}^{\infty}\varphi(t)dt=0$$
Then $$\int_{-\infty}^{\infty}\frac{|\hat{\varphi(\xi)}|^2}{\xi^2}d\xi<\infty$$
So far i have reached 
\begin{align*}
\int_{-\infty}^{\infty}\frac{|\int_{-\infty}^{\infty} \varphi(t)e^{-2\pi it\xi}dt|^2}{\xi^2}d\xi\leq\int_{-\infty}^{\infty}\frac{\int_{-\infty}^{\infty}|\varphi(t)|^2|e^{-2\pi it\xi}|^2dt}{\xi^2}d\xi\leq\int_{-\infty}^{\infty}\frac{1}{\xi^2}d\xi\int_{-\infty}^{\infty}|\varphi(t)|^2dt
\end{align*}
I am stucked here. Can someone help me? Also i am not really sure what it implies that a function has sufficient decay and vanishing mean. 
Thank you
 A: Hint: The vanishing mean condition
$$\int_{-\infty}^\infty \varphi(t)\:dt = 0$$
implies that there exists an antiderivative of $\varphi$ that belongs to the same "smooth functions with sufficient decay" space (here sufficient decay means assume as much regularity as you need to). Then use Plancherel's theorem.

$\textbf{EDIT}$: From the vanishing mean condition we get there exists $\psi$ such that
$$\psi'(x) = \varphi(x)$$
and 
$$\int_{-\infty}^\infty \psi(t)\:dt = M < \infty$$
From here we can use Fourier tranform properties to determine that
$$\hat{\psi}(\xi) = \frac{\hat{\varphi}(\xi)}{i\xi} + \hat{\varphi}(0)\delta(\xi) = \frac{\hat{\varphi}(\xi)}{i\xi}$$
since 
$$\hat{\varphi}(0) = \int_{-\infty}^\infty \varphi(x)e^{-i2\pi\cdot 0 \cdot x}\:dx = \int_{-\infty}^\infty \varphi(x)\:dx = 0$$
and we have that
$$\int_{-\infty}^\infty |\psi(x)|^2\:dx < \infty$$
due to the sufficient decay condition. Thus by Plancherel's theorem:
$$\int_{-\infty}^\infty |\psi(x)|^2\:dx = \int_{-\infty}^\infty |\hat{\psi}(\xi)|^2\:d\xi = \int_{-\infty}^\infty \frac{|\hat{\varphi}(\xi)|^2}{\xi^2}\:d\xi < \infty$$
